Question: Let $f(x) = -10x^{2}+10x-1$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Solution: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-10x^{2}+10x-1 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -10, b = 10, c = -1$ $ x = \dfrac{-10 \pm \sqrt{10^{2} - 4 \cdot -10 \cdot -1}}{2 \cdot -10}$ $ x = \dfrac{-10 \pm \sqrt{60}}{-20}$ $ x = \dfrac{-10 \pm 2\sqrt{15}}{-20}$ $x =\dfrac{-5 \pm \sqrt{15}}{-10}$